integral of (ln(xe^x))/x

解

\int \frac{ln ( x e ^{x} )}{x} d x

x + \frac{1}{2} ln^{2} (x) + ln (x) - ln ∣ x ∣ - 1 + C

解答ステップ

\int \frac{ln ( x e ^{x} )}{x} d x

部分積分を適用する

= \frac{1}{2} (x + 2 ln (x) - 2) - \int - \frac{x + 2 ln ( x ) - 2}{2 x} d x

\int - \frac{x + 2 ln ( x ) - 2}{2 x} d x = - \frac{1}{2} x - \frac{1}{2} ln^{2} (x) + ln ∣ x ∣

= \frac{1}{2} (x + 2 ln (x) - 2) - (- \frac{1}{2} x - \frac{1}{2} ln^{2} (x) + ln ∣ x ∣)

簡素化

\frac{1}{2} (x + 2 ln (x) - 2) - (- \frac{1}{2} x - \frac{1}{2} ln^{2} (x) + ln ∣ x ∣) : x + \frac{1}{2} ln^{2} (x) + ln (x) - ln ∣ x ∣ - 1

= x + \frac{1}{2} ln^{2} (x) + ln (x) - ln ∣ x ∣ - 1

定数を解答に追加する

= x + \frac{1}{2} ln^{2} (x) + ln (x) - ln ∣ x ∣ - 1 + C